Final answer:
The student's question contains some inconsistencies and does not provide a complete range for the variable k. Given the incomplete information, we can state that for k=0 and k=1, there are trivial solutions for x, y, and z. However, without additional information or clarification, providing a definitive answer is not possible.
Step-by-step explanation:
The student asks for the values of x, y, and z such that x³+y³+z³=k for each k from 1 to person to get it right gets a crown, however, this is either a typo or not a complete range provided for k. We are provided with multiple sets of values for x, y, and z, but not all of them clearly fit into the context of solving the equation x³+y³+z³=k. So let's consider the cases of k being 0 and 1:
- For k=0, a trivial solution is x=0, y=0, and z=0.
- For k=1, we can also have trivial solutions, such as x=1, y=0, and z=0.
However, without a complete description for the range of k, finding values for x, y, and z based on the given information is not possible. The answer options given in the question (A, B, C, D) do not align with the typical format of a question asking to solve for variables in an equation and thus cannot be addressed.
Cubing of Exponentials
We have reference to the cubing of exponentials, which means to raise each term with an exponent to the power of 3. This concept can be utilized to simplify the equation x³+y³+z³=k for certain values of x, y, and z when these variables are exponentials.